Question

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Short answer

Question: Prove that for any angle \(x\) such that \(0 < x < \frac{\pi}{2}\), the inequality \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\) holds. Answer: One way to prove this inequality is by comparing the areas of different figures. First, draw a unit circle and create the right triangle, isosceles triangle and sector with angle \(x\). Then, compare their areas, and it becomes clear that the right triangle's area is less than the sector's area, which is less than the isosceles triangle's area. Finally, using these area relationships, we can derive the given inequality \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\).

Step-by-step solution

Draw a unit circle and the angles

Draw a unit circle centered at the origin. Let \(O\) be the origin, \(A\) be the point on the positive side of the x-axis (on unit circle) where the angle starts, and \(B\) be the point on the positive side of the y-axis (on unit circle) where the angle ends. The angle between points \(A\) and \(B\) is \(x\) and we know that \(0 < x < \frac{\pi}{2}\).

Draw a right triangle, an isosceles triangle, and a sector

Connect point \(B\) to point \(A\) to form the right triangle \(OAB\). Also, draw a line segment from \(B\) to point \(C\) on the x-axis such that \(OAC\) is an isosceles triangle. Both triangles, \(OAB\) and \(OAC\), are formed within the sector \(AOC\) which has angle x.

Compare the areas

Now we will compare the areas of the different figures: 1. The area of the right triangle \(OAB\), denoted by \([OAB]\), is \(\frac{1}{2}OA . OB = \frac{1}{2}\sin x\). 2. The area of the sector \(AOC\), denoted by \([AOC]\), is \(\frac{1}{2}OA . OB . \angle{x} = \frac{1}{2}x\). 3. The area of the isosceles triangle \(OAC\), denoted by \([OAC]\), is \(\frac{1}{2}OC . AC = \frac{1}{2}(\sin x)(\tan x) = \frac{1}{2}\sin^2 x.\) Clearly, we have \([OAB] < [AOC] < [OAC]\).

Apply the inequality

Since \([OAB] < [AOC] < [OAC]\), we have $$\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\sin^2 x$$ Multiplying all sides by \(\frac{2}{\sin x}\), we get $$\frac{2}{\pi} < \frac{\sin x}{x} < 1$$ which is the inequality we set out to prove.

Key concepts

Unit Circle

The unit circle is a crucial concept in trigonometry, particularly when dealing with trigonometric inequalities. It is a circle with a radius of one, centered at the origin of the coordinate plane. Every point on the unit circle represents an angle in standard position, where the vertex is at the origin and one side is along the positive x-axis. This simple geometric figure helps to visualize the relationship between angles and trigonometric functions.
When we draw the unit circle for angles between 0 and \( \frac{\pi}{2} \), we focus on the first quadrant. Here, all trigonometric values are positive. The coordinates of any point on the unit circle, represented as \((\cos x, \sin x)\), relate directly to the angle \(x\). In this context, the sine of the angle \(x\) is the y-coordinate of the corresponding point on the circle.
Understanding the unit circle allows us to explore these relationships and prove inequalities involving the sine function, such as \( \frac{2}{\pi} < \frac{\sin x}{x} < 1 \).

Right Triangle

A right triangle is a triangle where one of its angles is exactly 90 degrees, or \(\frac{\pi}{2}\). In trigonometry, right triangles play a vital role in understanding the basis of trigonometric functions. Specifically, by connecting points on the unit circle, we can form right triangles that help to explore and establish relationships between angles and their trigonometric ratios.
In the proof of the inequality \( \frac{2}{\pi} < \frac{\sin x}{x} < 1 \), we consider a right triangle \( OAB \) on the unit circle, where \( A \) is on the x-axis and \( B \) corresponds to the point \((\cos x, \sin x)\). The hypotenuse \( OA \) is the radius of the unit circle, equal to 1, making it easier to compute the area of the triangle and use it in comparisons.
The area of triangle \( OAB \) is \( \frac{1}{2}\sin x \), illustrating its dependence on the sine of angle \(x\). By examining this and other areas within the circle’s sector, we gain insights into the relationships among the trigonometric functions.

Area Comparison

Area comparison is a powerful technique often employed in geometry and trigonometry to derive or prove formulas and inequalities. In this context, we explore the areas of different geometric figures related to the unit circle and a sector with angle \(x\).
The technique involves comparing the area of the right triangle \([OAB]\) and the sector \([AOC]\) of the circle. Triangle \([OAB]\) has an area \(\frac{1}{2} \sin x\), and sector \([AOC]\) has an area \(\frac{1}{2} x\). The insightful observation is that these two areas help establish an inequality. Additionally, we incorporate the area of another triangle \([OAC]\) to sandwich the area of the sector between two bounds.
This relationship tells us that \([OAB] < [AOC] < [OAC]\). By manipulating this inequality, we derive the broader inequality \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\). Such comparisons allow us to understand how geometric properties provide bounds for trigonometric functions.

Sine Function

The sine function is one of the principal trigonometric functions, fundamental in understanding periodic phenomena and angles. In a right triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the hypotenuse. However, in the context of the unit circle, the sine of angle \(x\) represents the y-coordinate of the corresponding point on the unit circle.
The inequality \(\frac{2}{\pi} < \frac{\sin x}{x} < 1\) showcases how the sine function behaves compared to the linear function \(x\). This relationship is crucial in calculus and analysis, reflecting the small-angle approximation, where \(\sin x\) is nearly equal to \(x\) when \(x\) is close to zero.
Understanding how the sine function compares to \(x\) within the specified range helps deepen knowledge about function approximation and periodicity, exemplifying the delicate balance between approximation and exact values.